# Sampling MCQ || Sampling in communication Engineering Questions and Answers

1. Calculate the Nyquist rate for sampling when a continuous-time signal is given by: x(t) = 5 cos 100πt +10 cos 200πt – 15 cos 300πt

1. 100 Hz
2. 150 Hz
3. 300 Hz
4. 600 Hz

Explanation

Minimum sampling rate to avoid aliasing f= 2fm= (Nyquist rate)

ωm = max {100π, 200π, 300π} = 300π

fm = 150 Hz = maximum frequency of signal

Sampling frequency or Nyquist rate; fs = 2 × 150 = 300 Hz

2. Consider two real valued signals, x(t) band-limited to [- 500 Hz, 500 Hz] and y(t) band-limited to [-1 kHz, 1 kHz]. For z(t) = x(t)⋅y(t), the Nyquist sampling frequency (in kHz) is ______.

1. 3 kHz
2. 5 kHz
3. 10 kHz
4. 15 kHz

Explanation

Convolution theorem states that the multiplication in time domain is equal to convolution in frequency domain, and vice-versa, i.e.

z(t) = x(t) ⋅ y(t) ⇔ Z(f) = X(f) * Y(f)

where X(f) and Y(f) are the Fourier transform of x(t) and y(t) respectively.

The Nyquist sampling rate is given as:

fs = 2 × fmax

fmax = Maximum frequency present at the input

Application:

Given x(t) is band-limited to [-500 Hz, 500 Hz]

y(t) is band-limited to [-1 kHz, 1 kHz]

z(t) = x(t)⋅y(t)

Since Z(f) = X(f) * Y(f), the output spectrum will be:

∴ The range of convolution in the frequency domain is [-1500 Hz, 1500 Hz]

So, the maximum frequency present in z(t) will be 1500 Hz.

The Nyquist rate will now be:

fs = (2 fmax) is 3000 Hz or 3 kHz

3. The spectrum of a bandpass signal spans from 20 kHz to 30 kHz. The signal can be recovered ideally from the sampled values when the sampling rate is at least:

1. 20 kHz
2. 60 kHz
3. 50 kHz
4. 40 kHz

Explanation

Sampling frequency of bandpass signal is given by:

${f_S} \ge \frac{{2{f_H}}}{k}$

Where

$k = \left[ {\frac{{{f_H}}}{{{f_H} – {f_L}}}} \right]$

Analysis:

fH = 30 kHz

fL = 20 kHz

fH – fL = 10 kHz

k = [30/3] = 3 (floor value is 3)

${f_S} \ge \frac{{2{f_H}}}{k} \ge 2\frac{{30}}{3} \ge 20{\rm{\;}}kHz$

So minimum sampling frequency is 20 kHz

4. If the sampling is carried out at a rate higher than twice the highest frequency of the original single (fmax), then it is possible to receive the original signal from the sampled signal by passing it through

1. A high-pass filter with the cut-off frequency equal to fmax
2. A low-pass filter with the cut-off frequency equal to fmax
3. A high-pass filter with the cut-off frequency greater than fmax
4. A low-pass filter with the cut-off frequency lesser than fmax

Answer.2. A low-pass filter with the cut-off frequency equal to fmax

Explanation

In the sampling process, if the sampling frequency is greater than the Nyquist frequency for the reconstruction of the original signal, LPF is used.

The sampling frequency can take three possible values with respect to spectrum width (2 ωm)

(i) ωs > 2ωm

(ii) ωs = 2 ωm

(iii) ωs < 2ωm

If the sampling is carried out at rate higher than twice the highest frequency of the original single (fmax), then it is possible to receive the original signal from the sampled signal by passing it through A low-pass filter with the cut-off frequency equal to fmax

When ωs = 2ωm

The spectrum of the sampled signal will be: We see for ωs > 2ωm, there is no overlap between the shifted spectrum of X(ω). Thus, as long as the sampling frequency ωs is greater than the twice the signal bandwidth (2 ω­m), x(t) can be recovered by passing the sampled signal xs(t) through an ideal or practical low pass filter having bandwidth between ωm and (ωs – ωm) rad/sec.

5. A system has a sampling rate of 50,000 samples per second. The maximum frequency of the signal it can acquire to reconstruct is

1. 25 kHz
2. 50 kHz
3. 100 kHz
4. 10 kHz

Explanation

Nyquist Sampling Theorem:

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

fs ≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

Application:

Given fs = 50 k

We can write:

50k ≥ 2 × fm

fm ≤ 25 k

∴ The maximum frequency of the signal it can acquire to reconstruct is 25 kHz

6. A band-limited signal is sampled at Nyquist rate, The signal can be recovered by passing the samples through

1. RC filter
2. envelope detector
3. PLL
4. ideal low pass filter with appropriate bandwidth

Answer.4. ideal low pass filter with appropriate bandwidth

Explanation

Nyquist Sampling Theorem:

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

fs ≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

If the low pass filter is not ideal, just sampling at the Nyquist frequency will not work we need to sample above the Nyquist frequency.

For example:

We have assumed spectrum x(t) is band limited to ωm.

Let, ωs > 2ωm

Let ωs = 3ωm

The spectrum of the sampled signal will be: We see for ωs > 2ωm, there is no overlap between the shifted spectrum of X(ω). Thus, as long as the sampling frequency ωs is greater than the twice the signal bandwidth (2 ω­m), x(t) can be recovered by passing the sampled signal xs(t) through an ideal or practical low pass filter having bandwidth between ωm and (ωs – ωm) rad/sec.

7. If the lower sideband overlaps the basebands, the distortion is called

1. Cross-Over distortion
2. Aliasing
3. Crosstalk
4. None of these

Explanation

If the lower sideband overlaps the basebands, the distortion is called Aliasing. The main reason for aliasing is undersampling i.e fs < 2fm.

8. What is the maximum bit rate of a noiseless channel with a bandwidth of 1000 Hz transmitting a signal with two signal levels?

1. 2000 bps
2. 3000 bps
3. 4000 bps
4. 6000 bps

Explanation

Given,

Bandwidth = 1000 Hz

L = 2

Hence, Maximum bit rate = 2 × bandwidth × log2 L = 2 × 1000 × log22 = 2000 bps

9. Consider a real-value based-band signal x(t), band limited to 10 kHz. The Nyquist rate for the signal y(t) = x(t) $x(1 + \frac{t}{2})$ is

1. 60 kHz
2. 30 kHz
3. 15 kHz
4. 20 kHz

Explanation

Given

x(t) = 10 kHz signal

Then,

$X\left( {\frac{t}{2} + 1} \right) \Rightarrow 5\:kHz$

$y\left( t \right) = x\left( t \right) \cdot x\left( {\frac{t}{2} + 1} \right)$

Nyquist rate of y(t) = 2 [frequency of y(t)]

= 2 × 15 kHz

= 30 kHz

10. A sinusoid of 10 kHz is sampled at 15 k samples/s. The resulting signal is passed through an ideal low pass filter (LPF) with a cut-off frequency of 25 kHz. The maximum frequency component at the output of the LPF (in kHz) is

1. 25 kHz
2. 20 kHz
3. 10 kHz
4. 40 kHz

Explanation

After sampling, frequencies present will be:  ± fm ± nfs

Where, n = 0, 1, 2, 3, 4, …

For n = 0, frequencies present = ±10 kHz

For n = 1, frequencies present = ± 10 k ± 15 k = ±25 kHz, ±5 kHz

After passing through the ideal LPF having a cut-off frequency of 25 kHz, the highest frequency component present at filter output = 25 kHz.

Therefore, the maximum frequency present at filter output is 25 kHz

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